r-比较两列，如果其中一列等于NA，则使用另一列更新它

Question

我有一个数据帧，看起来像这样：

# A tibble: 15 x 3
hm_variant_id           x    y
<chr>                <dbl> <dbl>        
1 10_64249655_C_T   0.0749 NA           
2 10_92361338_T_C  -0.0346 NA           
3 10_5224192_T_C   -0.0683 NA           
4 10_25993366_G_C   0.0231 NA           
5 10_20364_A_G      NA    1.9           
6 10_20665_TAAAC_T  0.0202 NA           
7 10_22591_G_T      NA    0.8           
8 10_23187_G_A     -0.250  NA           

我想循环浏览我的数据，比较x和y，并以一种方式相互更新：

if x = NA, then x = log(y)
if y = NA, then y = exp(x)

我的尝试是：

test2 <- test %>% mutate(new_or = exp(hm_beta))

但这创建了一个新列，并不能真正比较x和y。

test <- as.data.frame(test)
calculate <- function(data) {
x <- as.numeric(data["x"])
y <- as.numeric(data["y"])
if (is.na(x) == TRUE){
x <- log(y)
}
if (is.na(y) == TRUE){
y <- exp(x)
}
}
But I get this error:
Error in calculate_beta_ors(test) : 
'list' object cannot be coerced to type 'double'

在这里最干净的方法是什么？理想情况下，有tidyverse，但无论哪种方式都可以。

Answer 1

使用简单的ifelse:

transform(df, x = ifelse(is.na(x), log(y), x), 
y = ifelse(is.na(y), exp(x), y))

这也可以用dplyr函数类似地编写：

library(dplyr)
df %>%
mutate(x = if_else(is.na(x), log(y), x), 
y = if_else(is.na(y), exp(x), y))
#     hm_variant_id      x    y
#1  10_64249655_C_T  0.075 1.08
#2  10_92361338_T_C -0.035 0.97
#3   10_5224192_T_C -0.068 0.93
#4  10_25993366_G_C  0.023 1.02
#5     10_20364_A_G  0.642 1.90
#6 10_20665_TAAAC_T  0.020 1.02
#7     10_22591_G_T -0.223 0.80
#8     10_23187_G_A -0.250 0.78

数据

df <- structure(list(hm_variant_id = c("10_64249655_C_T", "10_92361338_T_C", 
"10_5224192_T_C", "10_25993366_G_C", "10_20364_A_G", "10_20665_TAAAC_T", 
"10_22591_G_T", "10_23187_G_A"), x = c(0.0749, -0.0346, -0.0683, 
0.0231, NA, 0.0202, NA, -0.25), y = c(NA, NA, NA, NA, 1.9, NA, 
0.8, NA)), class = "data.frame", row.names = c(NA, -8L))

Answer 2

使用ifels：

df$new_or <- ifelse(is.na(df$x) & !is.na(df$y), log(df$y),
ifelse(is.na(df$y) & !is.na(df$x), exp(df$x), NA))
df
#      hm_variant_id       x   y     new_or
# 1  10_64249655_C_T  0.0749  NA  1.0777764
# 2  10_92361338_T_C -0.0346  NA  0.9659917
# 3   10_5224192_T_C -0.0683  NA  0.9339802
# 4  10_25993366_G_C  0.0231  NA  1.0233689
# 5     10_20364_A_G      NA 1.9  0.6418539
# 6 10_20665_TAAAC_T  0.0202  NA  1.0204054
# 7     10_22591_G_T      NA 0.8 -0.2231436
# 8     10_23187_G_A -0.2500  NA  0.7788008

Answer 3

我们可以使用case_when

library(dplyr)
df %>%
mutate(x = case_when(is.na(x) ~ log(y), TRUE  ~ x), 
y = case_when(is.na(y) ~ exp(x), TRUE ~ y))